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A telescope has an objective lens of focal length $200 \mathrm{~cm}$ and an eye piece with focal length $2 \mathrm{~cm}$. If this telescope is used to see a 50 meter tall building at a distance of $2 \mathrm{~km}$, what is the height of the image of the building formed by the objective lens?
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2167 Upvotes
Verified Answer
The correct answer is:
$5 \mathrm{~cm}$
From the formula for convex lens,
$$
\begin{aligned}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
& \therefore \quad v=\frac{f \times u}{u-f}=\frac{200 \times 200 \times 10^3}{\left[200 \times 10^3-200\right]}=\frac{200 \times 10^3}{999} .
\end{aligned}
$$
Also, magnification,
$$
\begin{aligned}
m & =\left|\frac{v}{u}\right|=\left|\frac{I}{O}\right| \\
& =\frac{200 \times 10^3}{999 \times 200 \times 10^3}=\frac{I}{50 \times 100} \\
I & =\frac{5000}{999} \simeq 5 \mathrm{~cm} .
\end{aligned}
$$
$$
\begin{aligned}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
& \therefore \quad v=\frac{f \times u}{u-f}=\frac{200 \times 200 \times 10^3}{\left[200 \times 10^3-200\right]}=\frac{200 \times 10^3}{999} .
\end{aligned}
$$
Also, magnification,
$$
\begin{aligned}
m & =\left|\frac{v}{u}\right|=\left|\frac{I}{O}\right| \\
& =\frac{200 \times 10^3}{999 \times 200 \times 10^3}=\frac{I}{50 \times 100} \\
I & =\frac{5000}{999} \simeq 5 \mathrm{~cm} .
\end{aligned}
$$
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